I have an 80 g·cm motor with a rotational frequency of 15,000 rpm. I want to lift a weight of 2 kg at a speed of 0.5 m/s. How do I calculate the gear ratio required?


2 Answers 2


I have a 80gcm motor with a rpm of 15000.
I want to lift a weight of 2 kg at a speed of 0.5 m/s.
How do I go about calculating the gear ratio required for this?

Firstly - is it possible?

In particular, is there enough input power available for the desired output power?

To within about 2% a very handy formula applies - it can be derived conventionally and seen that several factors happen to cancel nicely.

Watts = kg x metres x RPM

80 gram∙cm = 0.080 kg x 0.01 m

So for input W = 0.080 kg x 0.01 m x 15000 = 12 Watts.
This is the maximum Wattage you can deliver if properly geared at 100% efficiency
(we should be so lucky).

Desired power = Force x distance per unit time
Watts = Joules/sec = mg∙d/s

= 2 kg x $g$ x 0.5 m/s = 2 x 9.8 x 0.5 = 9.8 Watts

So to work at all overall efficiency needs to be at least 9.8/12 or greater than about 82%.
That's potentially doable but also potentially difficult.

Now to the actual problem.

The following assumes that the output weight or force is taken from the end of a radius of the driven "gear". If output is instead taken from eg a windlass drum at lower diameter to the driven gear the ratios will be scaled based on the relative diameters. Ignore that for now.

Torque_in x RPM_in = Torque_out x RPM_out at 100% efficiency

or RPM_out = Torque_in x RPM_in / Torque_out at 100% efficiency


RPM out = 0.080 kg x 0.01 m x 15000 RPM / (2 kg x 0.5 m) = 12 RPM

So gear ratio = 15000/12 = 1250:1

The specification of not just output Torque but actual force (2 kg x $g$) constrains the actual output pulley size if output is taken off at the pulley radius.


If I understand correctly the problem is like this

enter image description here

The velocity of the load is $R\omega=0.5=R\underbrace{\frac{\pi}{30}\frac{15000}{n}}_\omega$

Solving for $n$ we get

$$n=1000 \pi R$$

  • $\begingroup$ i notice your solution is independent of the weight being lifted. I use the resultant ratio to get my result torque(say its 5Nm) does this mean that i can lift a weight of just below 5N while maintaining the same velocity?(assuming unit radius) $\endgroup$
    – lee wei
    Mar 10, 2015 at 10:54
  • 1
    $\begingroup$ My answer is just a simple kinetic analysis assuming the motor is able to pull the load. In order to accelerate the load the motor will have to produce an output torque greater than the load torque $\frac{m g R}{n}$. Solving $T-\frac{m g R}{n}>0$, one obtains $m<\frac{n T}{g R}$ $\endgroup$
    – Julian
    Mar 10, 2015 at 11:30

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